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So the question says

How many different binary numbers can be formed using 8 bits if: 1- In each number there are 3 adjacent ones 2- In each number there are exactly 6 ones (Edit: Each Part is different and solved separately)

I solved easily using logic but the thing is, we need to solve using permutation and combination rules I don't have any clue about the right answers

I got 6 for the first and 28 for the 2nd

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  • $\begingroup$ Oh, read those as two conditions of one problem. $\endgroup$
    – Thomas Andrews
    Aug 5 at 23:25
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    $\begingroup$ Your answer for $1$ implies the condition is that there are no other $1$s other than the three? $\endgroup$
    – Thomas Andrews
    Aug 5 at 23:27
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    $\begingroup$ What is the problem composer's intent, with respect to the first problem? Is 1-1-1-1-0-0-0-0 a satisfying sequence? What about 1-1-1-0-1-0-0-0? $\endgroup$
    – user2661923
    Aug 6 at 0:45
  • $\begingroup$ For part $(2)$ there are $\displaystyle \binom{8}{2} = 28~$ ways of selecting two positions out of the $(8)$ for the $0$'s to be placed. This assumes sampling without replacement, where the order of selection is deemed irrelevant. $\endgroup$
    – user2661923
    Aug 6 at 0:47
  • $\begingroup$ Oh I am sorry for the misunderstanding Each part is separated, and solved separately $\endgroup$
    – Eimon
    Aug 6 at 2:46

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